Suppose $R$ is a relation on $A$, and let $S$ be the reflexive closure of $R$. Prove that if $R$ is symmetric, then so is $S$.
S is the reflexive closure of $R$, which means that $$\forall x \in A (x, x) \in S$$
This is nice.
Now, I assume $R$ is symmetric, which means that $$\forall x, y \in A ((x, y) \in R \rightarrow (y, x) \in R)$$
This is also nice, but how do I go about to prove that also $S$ is symmetric? I can't really figure out how to do it just from the fact that $S$ is reflexive and $R$ is symmetric.
Intuitively, it seems correct, because $S$ should complete $R$ from the reflexivity point of view. If $R$ is symmetric, it seems in some way that also $S$ could be, but I don't know how to prove it.
For example, suppose $$R = \{(a, b), (b, a), (a, a)\}$$
it's symmetric (if I am not wrong), but not reflexive.
$S$ should complete the reflexivity of $R$:
$$S = \{(a, b), (b, a), (a, a), (b,b)\}$$
and we can see, in this example, that if $R$ is symmetric, also $S$, but this is not a proof.
